Verify that |r^{t}(s)|=1.Consider the helix represented investigation by the vector-valued function r(t)= < 2 cos t, 2 sin t, t >.

Question

Verify that

‖ r^{t} (s) ‖ = 1

.Consider the helix represented investigation by the vector-valued function

r (t) = < 2 \cos t, 2 \sin t, t >

.

Answer 1

Given:The function

\underset{―}{r (t) = < 2 \cos t, 2 \sin t, t >}

Proofs:The curve in terms of arc length is,

r (s) = 2 \cos (\frac{s}{\sqrt{5}}) i + 2 \sin (\frac{s}{\sqrt{5}}) j + \frac{s}{\sqrt{5}} k

.On differenting the vector-value function r(s), we get

r^{'} (s) = - \frac{2}{\sqrt{5}} \sin (\frac{s}{\sqrt{5}}) i + \frac{2}{\sqrt{5}} \cos (\frac{s}{\sqrt{5}}) j + \frac{1}{\sqrt{5}} k

From this calcute

‖ r^{'} (s) ‖

as

‖ r^{'} (s) ‖ = \sqrt{{(- \frac{2}{\sqrt{5}} \sin (\frac{s}{\sqrt{5}}))}^{2} + {(\frac{2}{\sqrt{5}} \cos (\frac{s}{\sqrt{5}}))}^{2} + {(\frac{1}{\sqrt{5}})}^{2}}

= \sqrt{\frac{4}{5} + \frac{1}{5}}

= \sqrt{\frac{5}{5}}

= 1

Hence, it is proved that

‖ r^{'} (s) ‖ = 1

Expert Answer